Question

If things really do have a dual wave-particle nature, then if the wave spreads, the probability of finding the particle should spread proportionally, independent of the degree of spreading, mass, speed, and even Planck’s constant. Imagine that a beam of particles of mass m and speedv, moving in the x direction, passes through a single slit of width w . Show that the angle ${\mathit{\theta }}_{\mathbf{1}}$at which the first diffraction minimum would be found ( $\mathit{n}\mathit{\lambda }\mathbf{=}\mathit{w}\mathit{s}\mathit{i}\mathit{n}\mathbf{}{\mathit{\theta }}_{\mathbf{n}}$, from physical optics) is proportional to the angle at which the particle would likely be deflected $\mathit{\theta }\mathbf{\cong }\frac{\mathbf{∆}{\mathbf{p}}_{\mathbf{y}}}{\mathbf{p}}$ , and that the proportionality factor is a pure number, independent of m, v, w and h . (Assume small angles: $\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{\cong }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{\cong }\mathit{\theta }$).

Short answer

It is shown that${\theta }_1=4\mathrm{\pi \theta }\Rightarrow {\mathrm{\theta }}_1\mathrm{\alpha \theta }$

Step-by-step solution

The diffraction minimum:

It is known that the diffraction minimum can be obtained at${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{w}}$

Required Proof:

Since, ${\theta }_1$is a very small angle, then

${\theta }_1\cong {\theta }_1$

Here,${\theta }_1$is the angle at which the first diffraction minimum is obtained. So,

$$\begin{gathered} {\theta }_1=\frac{\lambda }{w} \\ {\theta }_1=\frac{h}{mvw} \end{gathered}$$

Now, the angle at which the particle deflected is

$\theta =\cong \frac{∆p_y}{p}$ ….. (1)

Since, it is known that

$∆y∆p_y\approx \frac{\sout{h}}{2}$

And the momentum, p = mv

So, substitute the above value into the angle equation (1), and you have

$$\begin{gathered} \theta \approx \frac{\sout{h}}{(2∆y)\left(mv\right)} \\ \approx \frac{\sout{h}}{2wmv} \\ \approx \frac{h}{4\mathrm{\pi mvw}} \end{gathered}$$

Now, from the above calculations, you can say that ${\theta }_1=4\mathrm{\pi \theta }\Rightarrow {\mathrm{\theta }}_1\mathrm{\alpha \theta }$and the proportionality factor is a pure number equals $4\mathrm{\pi }$.

Hence, it is proved that ${\theta }_1\alpha \theta$and the proportionality factor is a pure number.